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aroub
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]

estudier
 2 years ago
Best ResponseYou've already chosen the best response.0Can't even see that, is it 10 th root?

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0can u pls give me the link...i want to delet my account..i can't find it anywhere

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1I don't know.. You have to contact a moderator

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0cmon moderator show up now...

phi
 2 years ago
Best ResponseYou've already chosen the best response.0The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1@akash_809 you cannot delete OS accounts

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1the simple reason is that\[x^{10}=(x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) \(x=\sqrt{x^2}\)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1so since\[\sqrt{(x)^2}=\sqrt{x^2}=x\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(x)^{10}}=x\](because x to any even power is positive, whether x>0 or x<0)

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1does that help you @aroub ?

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0@TuringTest root of negative number is not defined...it will result in iota value...so x wont be the answer i think

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1not necessarily, let's assume x<0...

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1If you assumed that, the question will be wrong.

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0yes it is...\[lets take an example......\sqrt{2} is \not defined unless you are taking imaginary numbers into account then u wl get +1.414 i\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0sry reply got messed up i was just saying that square root of 2 is not defined unless you consider imaginary numbers as wel..

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1Put imaginary numbers on a side for now..

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=x\]now we do it with the negative of x\[\sqrt[10]{(x)^4} \times \sqrt[10]{(x)^3} \times \sqrt[10]{(x)^3}=\sqrt[10]{(x)^4}\times\sqrt[10]{(x)^3\cdot (x)^3}\]\[=\sqrt[10]{(x)^{10}}=\sqrt[10]{x^{10}}=x\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is x

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1Okay take this as another example: \[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\] Here, a does not need an absolute value. Why? You can assume that a>0

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0yes you would have to assume a >0

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1let's just look at powers of a...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\](since we already knew a>0 now with the negative of a...

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0yea i m waiting for that negative values of a...what would it give

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1Aoh, so you have to try the negative and the positive?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt[4]{(a)^2}\times\sqrt[4]{a}\times\sqrt[4]{a}=\sqrt[4]{a^4}=a\neq a\]because a>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1sorry, I mean we took the negative of a...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1let me write that again, gotta be careful here

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0that's the problem...the thing you did here is called fallacy of maths...http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a\neq a\]since we assumed a<0

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(1)(1)}=\sqrt{1}\sqrt{1}=i.i=1 ....so u hv 1=1...and certainly \it is \not valid..\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1same thing with the earlier problem...

akash_809
 2 years ago
Best ResponseYou've already chosen the best response.0and 1=1....surely not right...

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1yes, but\[\sqrt{(1)^2}\cdot\sqrt{1}=\sqrt1\cdot\sqrt{1}\] is not the same as\[\sqrt{1}\cdot\sqrt{1}\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1notice we have a^2 in the above, so we know *that* part is positive

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.1this turned out to be more subtle than I bet you realized @aroub :)

aroub
 2 years ago
Best ResponseYou've already chosen the best response.1lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up .' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D
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